Problem: Multiply the following complex numbers: $({-2+3i}) \cdot ({2-5i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-2+3i}) \cdot ({2-5i}) = $ $ ({-2} \cdot {2}) + ({-2} \cdot {-5}i) + ({3}i \cdot {2}) + ({3}i \cdot {-5}i) $ Then simplify the terms: $ (-4) + (10i) + (6i) + (-15 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (10 + 6)i - 15i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (10 + 6)i - (-15) $ The result is simplified: $ (-4 + 15) + (16i) = 11+16i $